Vector calculus, + finding parametric equation

  • #1

Homework Statement


The line L
L1: x=3+2t, y=2t, z=t
Intersects the plane x+3y-z=-4 at a point P. Find a set of parametric equations for the line in the same plane that goes through P and is perpendicular to L.

Homework Equations


cross-product
r=r0+t(vector) this is to get in parametric form typically

The Attempt at a Solution


Well lets just say first I'm having trouble visually what this question is asking (is there a program where I can graph this type of stuff in 3 space?)

substituting the values of "L" into the plane equation and then solved for "t" which was -1. I then plugged those values into the parametric equations of the line to get the point of intersection P(1,2,-1)

Here is where I get confused and visually a bit shaky. I decided to use the normal direction vector, I'll call it
"n1" from the plane=<1,3,-1> and "n2" from the line =<2,2,1>
I take the cross product with the resultant vector being <-5,3,4>
So I then use the parametric equation which is: L2 => x=1-5t, y=-2+3t, z=-1+4t

I'm think I'm done at this point because now I have to sets of parametric equations. To check my answer I set
L1 and L2 equal to each other, solved for the variables using elimination and found the intersection point to be the same as I found earlier.

I have no idea if I'm right or wrong, any tips would be helpful! Please don't just tell me the answer, I would prefr someone to point out a mistake and let me figure out the rest, after all, I'm taking math to actually grasp it :)

Thanks in advance!
Cheers
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement


The line L
L1: x=3+2t, y=2t, z=t
Intersects the plane x+3y-z=-4 at a point P. Find a set of parametric equations for the line in the same plane that goes through P and is perpendicular to L.

Homework Equations


cross-product
r=r0+t(vector) this is to get in parametric form typically

The Attempt at a Solution


Well lets just say first I'm having trouble visually what this question is asking (is there a program where I can graph this type of stuff in 3 space?)

substituting the values of "L" into the plane equation and then solved for "t" which was -1. I then plugged those values into the parametric equations of the line to get the point of intersection P(1,2,-1)

Here is where I get confused and visually a bit shaky. I decided to use the normal direction vector, I'll call it
"n1" from the plane=<1,3,-1> and "n2" from the line =<2,2,1>
I take the cross product with the resultant vector being <-5,3,4>
So I then use the parametric equation which is: L2 => x=1-5t, y=-2+3t, z=-1+4t

I'm think I'm done at this point because now I have to sets of parametric equations. To check my answer I set
L1 and L2 equal to each other, solved for the variables using elimination and found the intersection point to be the same as I found earlier.

I have no idea if I'm right or wrong, any tips would be helpful! Please don't just tell me the answer, I would prefr someone to point out a mistake and let me figure out the rest, after all, I'm taking math to actually grasp it :)

Thanks in advance!
Cheers
The intersection point is <1,-2,-1> but that's just a typo. Looks ok otherwise.
 

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